Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $p \neq 0$. $k = \dfrac{-8}{3p^2 + 2p} \div \dfrac{4}{3(3p + 2)} $
Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{-8}{3p^2 + 2p} \times \dfrac{3(3p + 2)}{4} $ When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ -8 \times 3(3p + 2) } { (3p^2 + 2p) \times 4 } $ $ k = \dfrac {-8 \times 3(3p + 2)} {4 \times p(3p + 2)} $ $ k = \dfrac{-24(3p + 2)}{4p(3p + 2)} $ We can cancel the $3p + 2$ so long as $3p + 2 \neq 0$ Therefore $p \neq -\dfrac{2}{3}$ $k = \dfrac{-24 \cancel{(3p + 2})}{4p \cancel{(3p + 2)}} = -\dfrac{24}{4p} = -\dfrac{6}{p} $